help-49

[wintry bane]

But we have a constraint that their max value is 3

[solemn prism]

ok

[wintry bane]

So we will subtract all the cases that dont meet our criteria

Which are

(4,0,0,0) (0,4,0,0) (0,0,4,0) and (0,0,0,4)

Other than those, all the sets of (p,q,r,s) meet our constraints

So 35 - 4 = 31

In these many ways the rectangles can be arranged

[solemn prism]

Damn

Everything was thanks to that formula

And doing new variables

[wintry bane]

Haha yeah

[solemn prism]

Damn bro Thank you!

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[last slate]

how is it u-1 on the top

my u sub = x^2 + 1

(x^2 + 1) - 1 = x^3??

well, 1/2x

so (x^2 + 1) -1 /x = x^3??

[velvet compass]

du = 2x dx
x^3 = x^2 * x = 1/2(u-1)du

[alpine gyro]

$\int{\frac{x^3}{u}\cdot\frac{du}{2x}$

[grand pond]

Good
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[terse lodge]

because dx = du/(2x) gets rid of one of the x's on top

[last slate]

ty

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[grand pond]

Jeremy

[worn halo]

sqrt(1 + x^2) is called a hyperbola I think, but sqrt(1 + x^2) and sqrt(x^2) are actually quite similar for large values of x

and I guess sqrt(1 - x^2) is different, because there is a minus sign

so it isn't defined everywhere

[velvet compass]

focus on domain

for the 1st one

domain is [-1, 1]

and the equation of the unit circle is given by x^2 + y^2 = 1

hence y = sqrt(1 - x^2) is the upper part of the circle

for the 2nd one domain is set of real numbers

and it's conjugate hyperbola

its equation is x^2/a^2 - y^2/b^2 = -1, for a = b = 1 we have:
x^2 - y^2 = -1
y^2 = x^2 + 1
y = sqrt(x^2 + 1) us upper part of it

what transformation do you want to use to go from the 1st one to the 2nd

generally taking root of f(x) is not described as a type of transformation

In other words there is no universal behaviour we can define for such an operation

like e.g. f(x) + c is vertical translation up c units

no matter what f(x) is

[grand pond]

Jeremy

[velvet compass]

for the square root situation is different

[velvet compass]

it is as it is, in math little detail can make a difference as you can see

why is it absolute value?

that's simple

consider two cases

x >= 0

then sqrt(x^2) = x

x < 0

sqrt(x^2) = x (???)

no because notice that if x was e.g. -5 then we would get that sqrt(x^2) = -5, which is impossible, square root is always nonnegative

and hence sqrt((-5)^2) = sqrt(25) = 5 and not -5

sometimes this property is even define as the definition of abs value

sqrt(x^3)

is defined for x >= 0

and its graph is different also

,w plot sqrt(x^3)

[grand pond]

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[glad canyon]

I need help with this assignment, specifically the 1-norm part.

[glad canyon]

My work so far:

[floral junco]

#linear-algebra might have better luck here

[glad canyon]

I'll try there as well ^^ Thank you

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[glad canyon]

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[burnt sequoia]

is (x^2)(y^2)(z^2) = (xyz)^2

[burnt sequoia]

so the sqrt(x^2y^2z^2) = xyz?

why is this true

I don't see it intuitevely

[merry spade]

(xx)(yy) = (xy)(xy)

using associative and commutative

[last slate]

No, the equation sqrt(x2y2z^2) = xyz is not always true.

[burnt sequoia]

\

[last slate]

it would instead be sqrt(x2y2z^2) = |xyz|

[burnt sequoia]

this is what im trying to understand

[last slate]

making sure there is an absolute value

[burnt sequoia]

oh yeah

that makes sense

[last slate]

cant read the handwriting

[merry spade]

it says its positive

in this case it is true

[burnt sequoia]

ok I understand now, thank you

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[normal plover]

given that f = {(1,2), (4,6), (9,7), (10,4) and g = {(5,3),(4,2),(3,7),(1,9)] evaluate f(g^-1(1)

[normal plover]

what would ur steps be

Im confused because of the inverse

[cosmic elk]

start with the inner part

g^-1(1)

[normal plover]

okay

so is g^-1(1) equal to 9?

[cosmic elk]

uhh no

hmm this is weird

[normal plover]

yea im confused

[cosmic elk]

we should be looking for a coordinate with a y value of 1, but it doesn't exist

so perhaps it is undefined

[normal plover]

ohh

cause my sister put the answer f(1) = 2

and the teacher marked it wrong

[cosmic elk]

well that would only be true if g^-1(1) = 1 lol

[normal plover]

yeah

[cosmic elk]

so I assume the answer is undefined

[normal plover]

wait so if theres an inverse you still look at the y part of the coordinate?

waut no

why are we looking for a y value of 1

ohhh

because the inverse of g^-1 should be the y coordinate but its not in any of the coordinates given

[cosmic elk]

yeah I mean it's basically asking what is the x coordinate of the coordinate with y = 1

[normal plover]

that makes way more sense

[midnight plank]

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[ember haven]

do i do the top equation times 2y or the bottom equation times -y

[ember haven]

solving system using elimination

just wanna make sure im doing it right LOL

[final plume]

You can do anything you want to solve this system

Some methods are just less hectic than others

And minimize the risk for mistakes

[ember haven]

got it

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[drifting umbra]

Hello! I need help with some algabra problems.

[dense idol]

common difference?

[drifting umbra]

isn

their a easier way

for that first one

?

[dense idol]

by knowing what the sequence is?

[dense idol]

or do you want to subtract 6 like 20 times

[drifting umbra]

-6

-6x20

or smth

[dense idol]

19 times

[drifting umbra]

why 19

[dense idol]

because it's from 1st term to 20th term

so you only do 19 in between

[drifting umbra]

-120

oh

[little hare]

jarasabee

[drifting umbra]

?

isnt -120

[drifting umbra]

excuse me

could you check that?

[little hare]

jarasabee's theorem is on this link www.yellkey.com/single

[drifting umbra]

<@&268886789983436800>

[dense idol]

no

[drifting umbra]

adverstiting

to a unkown link

[little hare]

click it

[drifting umbra]

could any one help

[little hare]

ok chill lol

its a yt channel

[drifting umbra]

<@&268886789983436800>

[little hare]

k mb

chill

[dusk furnace]

Huh, what?

[little hare]

ill delete it

[drifting umbra]

adversting unkown link

[little hare]

i said this as a joke

its a yt channel link lol

[drifting umbra]

he is intruping

our session

and telling us

to click a random link

[little hare]

ong its a yt channel

[drifting umbra]

for no reason

[dusk furnace]

Ah, there. Discord didn't even make the link clickable, so I missed it the first times, sorry.

[drifting umbra]

your fine

could you deal with the siuation

[little hare]

yo am i in trouble

[drifting umbra]

plz, kick or just talk to him

to him

either one is fine

just need help

[little hare]

ur mean

@dusk furnace sry but pls dont kick me at least

[dusk furnace]

It doesn't look like you're contributing to the help here, so please stop posting, or you will be.

[drifting umbra]

use a virus snaccer

[drifting umbra]

look at a ip link

[little hare]

ong its a yt channel

[drifting umbra]

scanner

[little hare]

test it if u want lol

[dense idol]

hackerman

[little hare]

it goes to this link

[drifting umbra]

and adversting is not allowed

<@&268886789983436800>

[little hare]

Oh

[drifting umbra]

he is adversting again

sorry for the ping

[dusk furnace]

Muted, since nice please to stop posting didn't work.

[drifting umbra]

ty

[drifting umbra]

sir, if you are not to busy could you help me?

[drifting umbra]

if you are back could you help

[dusk furnace]

I don't have time now, sorry.

[midnight plank]

@drifting umbra Has your question been resolved?

[midnight plank]

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[drifting umbra]

Hello

[drifting umbra]

Here

I'm not sure if you could see some of the pictures I had from the other one

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[drifting umbra]

could you help me

[midnight plank]

@drifting umbra Has your question been resolved?

[vivid lance]

@drifting umbra do you still need help?

sorry i was late

🥲

?

[drifting umbra]

Ok

@vivid lance

[vivid lance]

do you need help on the above stuff though

[drifting umbra]

Nah

I figured it out

[vivid lance]

Okay

[drifting umbra]

I need help with this one

The last one not the four

@vivid lance

[vivid lance]

what?

im very confused sorry

Oh i get it sorry

okay

ive never learned this before but i think the first one's equation would be a1+5d=39

because since we're seperating the a6

it would be 5d for another variable unless im not using my brain

then for a12 it would be a1+11d=81 i would assume

so then you would subtract 11d-5d=81-39

6d=42

which means d would equal 7?

so now time to find d

a1+5(7)=39? solve this for the first term

this means the difference is 7

i guess even though i haven't seen this stuff before

[drifting umbra]

Can you make that clear?

@vivid lance

I think that's correct

Can you just make it clear so I can understand

[vivid lance]

yes of course

okay so a6 = a1 + 5d = 39

d = difference

a1+5 is 6 so that adds up to a6

so it would equal 39

so then

for a12 it would be a1+11d=81 because 11+1 is 12

and you would subtract the equations to FIND the difference

so a1-a1 is 0

11d-5d=81-39

or we could do

5d-11d=39-81

which is the same thing

which means d=7

because -6d=-42

d=-42/6

oop

d=42/6 which equals to 7

then u subsitute d for the FIRST term

the first term IS a1+5d=39

so yeah it'd be a1+5(7)=39

then a1+35=39 then subtract 39

35*

then you'd get 4

for the first term

and the difference would be 7

🙂

im gonna get some sleep bye and good luck

[drifting umbra]

I got it. Thank you bro

. close

.close

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[somber patio]

can someone explain to me number 5 or the thing underlined in white

[somber patio]

I usually plug in numbers

But this will take a decade

[midnight plank]

@somber patio Has your question been resolved?

[last slate]

yo guys

what uni level r u at

im in grade 7

[midnight plank]

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[south zodiac]

Need help plz

Anyone

?

.close

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[keen zephyr]

I know the derivative of the function is 4x but I am unsure how to find the equation for the parallel line

for finding the point where the slope is parallel I got (1,2)

CNX^n-1

[balmy brook]

yes sorry

i'm dumb

[keen zephyr]

then point slope form would be y-2=4(x-1) right

y=4x-2?

did I solve that right?

<@&286206848099549185>

[midnight plank]

@keen zephyr Has your question been resolved?

[keen zephyr]

<@&286206848099549185>

[stray latch]

You solved that right

[keen zephyr]

Awesome thank you I only had one submission left so had to make sure

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[latent seal]

how do we go from the top to the bottom

OH

IM STUPID

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[vernal void]

Need help on this question

[midnight plank]

@vernal void Has your question been resolved?

[dense idol]

sample mean: $\overline {x}$

[grand pond]

chlamydia

[dense idol]

population mean: $\mu$

[grand pond]

chlamydia

[vernal void]

Not yet

[dense idol]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[midnight plank]

@vernal void Has your question been resolved?

[vernal void]

A. X is supposed to equal whatever number close and B that have that sign that looks like a U = something else

[midnight plank]

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[velvet parrot]

need help figuring out what I did wrong

[velvet parrot]

This is what i did

[velvet parrot]

.close

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[glass bloom]

Why is -(3^2)^3=-729 and not 729

[glass bloom]

Ik 3^2 is 9

Then do I make it negative then do the power or do the power then make it negative

[prisma idol]

cause -ive to odd power = -ive

[clever sedge]

Think of it as $(-1)^3 * (9)^3$

[grand pond]

USS-Enterprise

[flat spire]

but also the - is on the outside so even if you had -(3^2) it would still be -9

[clever sedge]

(-1)*(-1)*(-1) = 1*(-1) = -1

[glass bloom]

Im kinda confused

[clever sedge]

Yes

[glass bloom]

Wait so since the bracket is 9 would I do 9 times -1 then to the power of 3 or would I cube it first then multiply by negative one

[clever sedge]

Nevermind, forget what I said completely

I thought the minus was in parenthesis

[glass bloom]

Oh alr

[clever sedge]

You've got $-(9)^3$

[grand pond]

USS-Enterprise

[glass bloom]

Yeah

[clever sedge]

You can write this as $(-1) * (9)^3$

[grand pond]

USS-Enterprise

[clever sedge]

So no matter what comes out in the second one it's getting multiplied by (-1)

[glass bloom]

Ah I see so I'd cube it first

[clever sedge]

So it will always be negative, even if the power is even

Yes

[glass bloom]

Thanks I get it

[clever sedge]

Yeah, sorry I misread the original problem

[glass bloom]

Np

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[sharp ruin]

can anybody explain how you would go from y = x^2, to y = (1/16)(x+2)^2+1

[sharp ruin]

I wrote vertical compression of 1/16, translate to the left by 2 units, upwards 1 unit.

but apparently it's: stretch horizontally by a factor of 4, translate left by 2 units, upwards by 1

[midnight plank]

@sharp ruin Has your question been resolved?

[sharp ruin]

<@&286206848099549185>

[midnight plank]

@sharp ruin Has your question been resolved?

[dense latch]

So you mean you can't understand this?

[sharp ruin]

yeah

how is it being stretched horizontally by 4?

doesnt the 1/16 mean it's a vertical compression of 1/16?

where is the 4, horizontal, and stretch part coming from?

[dense latch]

Do yoi know how to stretch horizontally?

Actually I think vetically by 1/16 is same as horizontally as 4

In this case I mean

[sharp ruin]

oh

[sharp ruin]

y = ((x/2)+2)^2+1

?

[dense latch]

by 4

[sharp ruin]

ye

[dense latch]

$y=(\frac{x}{4}+2)^2+1$

[sharp ruin]

yeah ik

mistyped

how does that equate to (1/16)(x+2)^2+1 tho

[dense latch]

i miswrote

$y=(\frac{x+2}{4})^2+1$

[grand pond]

Dri111

[dense latch]

because we first stretched, then moved

[sharp ruin]

o

ok thx

appreciate it

.close

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[main geyser]

Hello I need help with this question

[main geyser]

So for (A) I think its 462 elements since there are 6 spots to fill, we can choose it in (11 choose 6) different ways, so our sample space S will contain 462 elements

This is because the prime factors of the above number are P = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 }

For (B), can I just remove 3 from the factors list and then just find that P(A) = 1-P(not A) = 1 - (10 choose 6/11 choose 6) = 6/11

(C) is apparently a typo in question and its suppose to be "product is divisible by 6". So im kinda confused how to go on this, the prime factors of 6 is 2 and 3 so does that mean i just remove 2 and 3 from the P again and do something similar???

So for (C) it would be 1-[P(not divisible by 3) + P(not divisible by 2) - P(not divisby by 2 and 3)] = 1-[(10 choose 6/11 choose 6) + (10 choose 6/11 choose 6) - (9 choose 6)/(11 choose 6)] = 3/11?

I dont really know how to do (E)...... Can i get some help on this one as well

anyone?

<@&286206848099549185>

Actually for (E), if ordered matter, then therer would have been 11 choices for first position, 10 choices for 2nd position and etc until 6th position. So total elements would have been 11*10*9*8*7*6=332640. Since there is only one unique way the subset can be increasing order, this is just the same as asking how many ways can we find a 6 digit number. So this is just (11 choose 6)/332640 which is 1/720?

is this correct????

[midnight plank]

@main geyser Has your question been resolved?

[main geyser]

no halp?

[midnight plank]

@main geyser Has your question been resolved?

[midnight plank]

@main geyser Has your question been resolved?

[midnight plank]

@main geyser Has your question been resolved?

[midnight plank]

@main geyser Has your question been resolved?

[vague seal]

So for B what they are asking is what is the probability that in the 6 you've chosen out of 11 one of them is 3

And for C it is what is the probability that both 2 and 3 are in there

[main geyser]

so if i take the compliment of it, P(not divisible by 3), it would not include 3 at all, so it would be (10 choose 6) / (11 choose 6)

so this results in P(divisble by 3) = 6/11, is this correct?

[midnight plank]

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[midnight plank]

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[weak summit]

I am attempting to prove that for the functions g(n) and f(n) have the same asymptotic growth rate. The exponent of g(n) is a constant 3/2. The exponent of f(n) is periodic between 1-3 is it correct to say that at infinity/large enough valaues these two functions will have the same growth rate?

[weak summit]

For the sake of this example. Let us say that f(n) = x^(1 - 3) and g(n) = x^3/2

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[steep nimbus]

can someone help me with this

[neon rune]

Well, there is a really dirty trick

Consider the number closest to it
Let it be x
Now divide the square of that number by x
After which you take their average value.
It seems like out of nowhere but it is a very close approximation
REMEMBER to use the nearest number

But since maybe you are in school, you would be forced to do it the 'school' method AKA actually finding the square root by long division

[steep nimbus]

yes

[neon rune]

Do you know how to find square roots by long division?

[steep nimbus]

yes

[neon rune]

Then you should proceed with it

[steep nimbus]

so i find the square root of 53 and i convert to 1 decimal place?

[neon rune]

just find the first two places

[last slate]

Just find the two decimal places, round it off to one decimal place

[steep nimbus]

can someone tell me the straight up answer?

[neon rune]

sadly no

[steep nimbus]

its fine i did it 10 minutes ago

[midnight plank]

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[livid bramble]

could someone guide me

[stark hazel]

i think you will need to use the fact that f is uniformly continuous

since its on a compact interval

[gusty shuttle]

uniform implies pointwise. find f by taking fn(x) n to ifty limit.
Now u have candidate f for convergence.
Take that f and proove uniform convergence

[stark hazel]

well, f is given in the problem. so they don't have to construct it

which makes things easier

[gusty shuttle]

oh right my bad

but only implicitly

[stark hazel]

my hint is to use/think about uniform continuity. but i can help/provide more if needed

[gusty shuttle]

gotta find the f(x) expression without n in argument

then u have true f

[stark hazel]

it starts out with "Let f b e a continuous function on [0,1]". so you have f already given to you. Arkos needs to prove that f_n --> f uniformly, where f_n is (conveniently) defined in terms of f

[gusty shuttle]

i see

[midnight plank]

@livid bramble Has your question been resolved?

[livid bramble]

\forall \epsilon \exists \delta \forall x,y (|x-y|<\delta => |f(x)-f(y)|)
y:=(k-1)/n
for that we need 0<=x-(k-1)/n<1/n

bit hazy here

so set N=floor(1/delta)?

[stark hazel]

yeah. that's right.

then you can check sup_{x \in [0,1]} |f_N(x) - f(x)| \leq \epsilon

as well as for all n >= N

[livid bramble]

im still bit blurred at how its working
if N=floor(1/delta)
delta>1/n and delta>|x-(k-1)/n|
but I can't seem to be getting a hold of 1/n>|x-(k-1)/n| which I need for k to be defined

[stark hazel]

what do you mean that you need for k to be defined?

i was typing the next step but it occurred to me that talking about the confusion might help more

[livid bramble]

like k depends on x and n
and for f_n(x)=f((k-1)/n) we need 0<=x-((k-1)/n)<1/n

[stark hazel]

oh, you only need to show that |x - (k-1) / n| < \delta, not < 1/n

[stark hazel]

yeah

[livid bramble]

so I really need x-(k-1)/n<1/n for k to be well defined

or we have to tweak it ig

maybe its all gibberish that im saying

[stark hazel]

maybe it helps to say it like this. the background: you have some \epsilon > 0, and you chose n \in N large enough such that, whenever |x - y| < 1 /N, you have |f(x) - f(y)| < \epsilon.

so now, if x is in [(k-1) / N, k /N) for some k, then |x - (k - 1) / N| < 1 / N. this means that |f((k - 1) / N) - f(x)| < \epsilon for all x in this half open interval

this leaves out the case when x = 1, but that's no problem. so this shows that |f_N(x) - f(x)| < \epsilon for all x \in [0,1] (because the half open intervals partition this set), hence the sup is <= \epsilon, so ||f_N(x) - f(x)|| <= \epsilon, where the norm is the supremum norm

now you have to show it for all n >= N

but this is no problem, its really just the same argument, since 1/n < 1/N

so you just have to check the inequality for all the x in each half open interval. i think that's what you were confused about. each time you pick an x, it lies in some half open interval

and whenever it does, |f_n(x) - f(x)| < \epsilon

maybe i could have done a better job explaining i haven't been doing this (helping in math channels) for that long

[stark hazel]

i think it was a little strange. you're saying |x - (k -1) / n| must be smaller than 1 / n for k to be well defined. but your condition for requiring k to be well defined itself uses k

[livid bramble]

idk at this point
f_n(x)=f((k-1)/n), where k is a positive integer that satisfies these inequalities 0<=(x-(k-1)/n)<1/n

[stark hazel]

yeah. so $f_n$ is defined piecewise by $$f_n(x) = \begin{cases} f((k-1) / n) & x \in [(k-1) / n, k / n)\ f(1) & x = 1 \end{cases} $$

jesus

[livid bramble]

I feel like I should be able to do it now, I'll comeback if I get stuck

[grand pond]

ves

[stark hazel]

okay good luck. it helps maybe to try to picture the function f_n

[livid bramble]

cool, after writing it down
I understand it well

thanks a lot

.close

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[midnight plank]

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[honest shuttle]

How do you do conjuctive normal form from logical expression to boolean without using K-map?

[honest shuttle]

I mean if we get a disjunctive form, how do I convert it to conjunctive?

For example, this what I got in DNF from one function and I can easily make a conjunctive if I take K-map into consideration

But I need to show some steps on how I calculated CNF like in DNF.

Yet I have no clue how or where to start with CNF

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[marsh marsh]

can someone help me understand how they got -4?

[marsh marsh]

i dont even know how to start

[subtle blaze]

There’s a typo

They mean where k is a constant

Not where f is a constant

Does that help?

[marsh marsh]

no i still dont know how to do the problem

[tight compass]

Set equal and solve for k as it is continuous.

[marsh marsh]

i tried conjugation but i dont think i was supposed to

[subtle blaze]

What does it mean for f to be continuous at x = 0?

[marsh marsh]

wym set equal

[subtle blaze]

Don’t spoil it

@tight compass

[tight compass]

ok.

[marsh marsh]

???

[subtle blaze]

Thanks :p

[subtle blaze]

@marsh marsh answer this

[marsh marsh]

that means f is defined at x=0 right

[subtle blaze]

That’s a necessary but not sufficient condition

[tight compass]

What three conditions must be true for a function to be continuous?

[marsh marsh]

it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point

[subtle blaze]

Can you more precisely define what you mean by “the limit must exist at the point”

[marsh marsh]

idk how

[subtle blaze]

Have you seen left and right hand side limits

[marsh marsh]

yea

[subtle blaze]

That

Use those terms to more precisely define what you mean by the limit “existing”

[marsh marsh]

left and right hand limits are equal

[subtle blaze]

Aha

[subtle blaze]

Find a k that makes this true

[subtle blaze]

And this true

Then f will be continuous at x = 0

[marsh marsh]

well i know the answer is -4 but i just dont know the process of getting to it

[subtle blaze]

Well let’s check each thing

What’s the first condition

[marsh marsh]

must be defined at x=0

[subtle blaze]

Is f defined at x = 0?

[marsh marsh]

yes

[subtle blaze]

Great

First condition done

What’s the next condition

[marsh marsh]

limit must exist at x=0

[subtle blaze]

What does that mean

[marsh marsh]

left and right hand have to be equal

[subtle blaze]

Ok

What’s the left hand limit

[marsh marsh]

positive infinity

wait no

what am i supposed to substitute x with

4?

[subtle blaze]

We want to find $\lim_{x\to 0^-} f(x)$ right?

[grand pond]

Frosst

[subtle blaze]

Hold up why does it say “and” in the first line of the piecewise function

[marsh marsh]

yea

idk

[subtle blaze]

x ≠ 0 and x >= 4 just means x >= 4

Unless it really meant “or”

[marsh marsh]

yea

[subtle blaze]

But x ≠ 0 or x >= 4 is just x ≠ 0

This is really weird

Well I’m gonna assume they meant or

Because if they meant and, then f is not continuous at x = 0 for any k

Because your function’s domain would just be {0} ∪ [4, inf)

[tight compass]

(-4, inf)

[subtle blaze]

Why -4

[tight compass]

well, the domain of the function is x>-4

everything greater than.

[marsh marsh]

isnt it greater than or equal to

[subtle blaze]

Yeah but look at the conditions on the piecewise

If they really meant “and” then it won’t include anything less than 4 except 0

Hence I will assume they meant “or”

[tight compass]

x>=4 is useless to the function, i did not bother with it 🤷‍♂️

[marsh marsh]

what did you do then

[tight compass]

if you plug in 0 for the top function, you get 0/0

what is the rule if you get 0/0? what must you do?

[marsh marsh]

try to change the function to get a different answer

like factor

or conjugation

[tight compass]

factor conjugate yes!

[marsh marsh]

cuz indeterminate form

yes

[tight compass]

tell me your factored form.

[marsh marsh]

i tried conjugation but i either did it wrong or it doesnt work for this

[tight compass]

what is the conjugate of $2-\sqrt(4+x))$

[marsh marsh]

2+sqrt(4+x)

[grand pond]

Tank_Driver011

[tight compass]

yes.

what do you do with this?

[marsh marsh]

multiply to top and bottom

[tight compass]

what happens when this is evaluated? what do you get?

[marsh marsh]

hold on let me draw it

i get this

[tight compass]

yes.

now simplify (factor negative sign out)

[marsh marsh]

and then when i plug in 0 i get -4

right?

[tight compass]

ok good! your professor is going to take points off from this, there is just one crucial thing you must write on the side of your shown work relating to the limit: What is it?

[marsh marsh]

am i supposed to make it g(x)?

cuz its supposed to be like im using an identical function

[tight compass]

you are supposed to show the limit (what is happening)

[marsh marsh]

so i write lim x->0?

[tight compass]

Yes! Nice work!

plug in 0 and you get your value.

the function does not show a split, we know it is continuous.

if it had x>0 x<0 that would be a different case to check if it checks out with all of the rules.

now you have k solved!

@subtle blaze good? Anything you wanted to add?

[marsh marsh]

alr ty 🙏🏼

[subtle blaze]

Yeah that’s good

[marsh marsh]

how do you know the function is continuous

[subtle blaze]

You want to show all the conditions of continuity is held by your k

[subtle blaze]

You said it yourself

[subtle blaze]

This

[subtle blaze]

And this

If bla bla bla is true, we call it “continuous”

To show continuity you show that bla bla bla is true

[tight compass]

K is a constant, we know it checks out, it tells us this.

the right and left side are going to 0, they must as it was solved for (it is a 0/0 problem, we use lim x approaches n, not stating right and left sides) in the top function which equated to -4. f(x) = -4, and lim x approaches -4 = f(x) = -4

could be written as lim x approaches -4 = f(0)=-4 as well

[marsh marsh]

ah ok

ty both 🙏🏼

.close

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[last slate]

If the farmer has 184 feet of fencing, what is the largest area the farmer can enclose?

[last slate]

umm

How do i do that

:conversion

<@&286206848099549185>

[midnight plank]

@last slate Has your question been resolved?

[tight compass]

set your equation solving for perimeter with variables L (length) and W (width) first.

[tight compass]

Just realized what you were saying here, you are right. I was speaking of the function without the restrictions. Sorry about that. 😄

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[last slate]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[last slate]

2!!!

$[0,1]$ is in $\mathbb{R}$ and ${0}$ is in $\mathbb{R}$

[grand pond]

What should I do: Physics

[last slate]

BUT

The probability of ${n}$ where $n \in \mathbb{R} = 0$ right?

[grand pond]

What should I do: Physics

[last slate]

$P({n}) = 0$ , $n \in \mathbb{R}$ ?

[grand pond]

What should I do: Physics

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

I’m so confused on these problems

I don’t understand

[midnight plank]

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[last slate]

Pls help

!msgdel

[midnight plank]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[last slate]

Why

Bruh

Can I get help pls

claim a new channel, this channel will lock abruptly

[last slate]

...

.reopen

.open

.close

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[last slate]

There’s no

[last slate]

Help please. Idk how to do this

[oak pollen]

You just have to take limit and see which function will go to negative infinity. So answer should be 1st, 3rd, 5th options.

Let us take 1st for example.

[last slate]

It’s only the first and second

Not fifth

I gotta figure out why tho

[oak pollen]

Yeye, sry.

I thought it was sth else.

Ok, so 1st and 3rd.

[last slate]

Yea

How thi

Tho

Here’s another one

[oak pollen]

As x approaches more and more negative value, the value of f(x) will approach negative infinity or positive infinity depending on leading coefficient sign at x equals something negative. Just check using -1. If leading term gets a positive coefficient, it is positive infinity. Or else negative infinity.

[last slate]

Ok

This one is first second and fourth right?

@oak pollen

[oak pollen]

Just a second.

1st 2nd and 4th.

Yes.

[last slate]

I just guessed that cause it’s positive and the rest are negative

Cause the one in parenthesis had a -3 which would make it a -Infiniti right?

[oak pollen]

Yes.

-infinity at both x tends + and - infinity.

So decreasing.

[last slate]

Ok thanks for the help

[oak pollen]

You are welcome.

[last slate]

.close

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[weary swallow]

can someone help me chose what classes to have for math in high school for an 8th grader and in what order? (we can go to dms)

[weary swallow]

any <@&286206848099549185> here that can help?

[tulip token]

honestly take whatever you find interesting

if you don't have a choice, and if you're asking the usual progression of classes, then it goes something like algebra 1, geometry, algebra 2, precalculus, calculus

[weary swallow]

i am currently doing algebra 1 in 8th grade

[tulip token]

but in any case, most of those classes are just review

[weary swallow]

also, i would like to prepare for geometry (things that arent covered in algebra 1)

<@&286206848099549185> ?

tldr; i want to practice geometry and learn about it

[tulip token]

maybe try some online geometry exercises

[midnight plank]

@weary swallow Has your question been resolved?

[tulip token]

or just find a high school geometry textbook / video series and read ahead

[midnight plank]

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[midnight plank]

@dusty rover Has your question been resolved?

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[spiral tangle]

Any short way?

[nocturne jetty]

?

that's the short way only

the d term gets cancelled

you get 'a'

[spiral tangle]

Thanks

.close

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[stone sky]

Show that the class of residues {2,3,...,p-2}=A can be paired up into multiplactive inverses, p prime.

[stone sky]

I have a weird proof, but im wondering how is this supposed to be done profesionally?

[runic hamlet]

well really the only thing you have to check is that there are no self-inverse elements

otherwise if a is not self inverse and b is its inverse, then the inverse of b is again a and you have the pair (a,b)

[stone sky]

righttt since the inverse is always in Zp?

[runic hamlet]

yes

cant be 0

and 1 and p-1 are self inverse, so cant be those

[stone sky]

obiously, Yeah that makes sense

Can i prove the existence of an inverse by saying that for some $a \in A$ let $x$ be the inverse

[grand pond]

bigpufik

[stone sky]

and let x not be in A

[runic hamlet]

oh you havent done existence yet?

[stone sky]

Im just trying to be sure... I am self studying and its all very icey

I assumed the above, and then claimed that $a\equiv r \pmod p$ where r, r' is a part of Zp and $x \equiv r' \pmod p$

[grand pond]

bigpufik

[stone sky]

then that means $rr' \equiv 1 \pmod p$

[grand pond]

bigpufik

[stone sky]

and so if the inverse wouldnt be in Zp, then it would be false.

is this ok?

I already proved the existence of the inverse using Zp and Zp where you multiply everything by a

[runic hamlet]

why not just directly write xa=1 mod p

that's the definition of what inverse means

[runic hamlet]

if you have proven existence with this then you are done

[stone sky]

but is this ok?

[runic hamlet]

you have proven the existence in Zp. therefore where else should it be

it makes no sense to talk about an inverse if that inverse is not in the same set

[stone sky]

right.

ok thanks

.close

[midnight plank]

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[shell oasis]

Y=MX+B solve for b

[mellow sand]

Feel like you are missing something

[shell oasis]

like hat

what

[mellow sand]

Then isolate B?

[shell oasis]

this is literal equasions

[mellow sand]

Ok

I was wondering if

Do you know how to isolate B?

[shell oasis]

divide m by x

[mellow sand]

That would make y/x = m + b/x

Is there something easier to do

[shell oasis]

idk

in not very sure

[mellow sand]

What happens if you subtract MX

[shell oasis]

wouldnt they cancel out?

[mellow sand]

Wdym?

[shell oasis]

idk anything about this type of math

[mellow sand]

All you need to do is isolate B

you can subtract M times X from one side of an equation

[shell oasis]

i still dont understan

[mellow sand]

Y = MX + B
Y - MX = B

[shell oasis]

ok thanks

.close

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[queen thistle]

((𝑛/𝑝𝑟𝑖𝑜𝑟𝑎𝑝𝑝𝑟𝑜𝑥)+𝑝𝑟𝑖𝑜𝑟𝑎𝑝𝑝𝑟𝑜𝑥)/
2

[queen thistle]

So I got the basic idea that this is supposed to approximate the square root using a guess

But I'm not exactly sure like what the answer is

So if I say 16 and my guess is 3

16/3 + 3 all divided by 2

4.166666667

I'm just not sure how I'm supposed to interpret this

[midnight plank]

@queen thistle Has your question been resolved?

[midnight plank]

@queen thistle Has your question been resolved?

[midnight plank]

@queen thistle Has your question been resolved?

[midnight plank]

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[midnight plank]

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[keen seal]

https://cdn.discordapp.com/attachments/789649003188715540/1155998112867942580/IMG_8499.jpg?ex=65175323&is=651601a3&hm=bda5dbe449a5f36c4b3a169492a28d0c967f25854ce5660fef94016071380f46&

[keen seal]

question 38

the answer is 0 and -1

but how is it possible to get that?

I get stuck after making this far

[amber wedge]

That's not the correct way to do this

[opal moat]

Isnt he curious why they balance

[amber wedge]

To write the full solution, you will need to prove that there is no other x for which f(x)=f(f(x))

[opal moat]

All u man ur probably better

[amber wedge]

Your way to do this is correct, but you will need to multiply both the numerator and the denominator by (x+2)

After simplify f(f(x)), calculate the equation f(f(x))=f(x)

[keen seal]

So ill assume ill have to turn the +2 into 2(x+2)/x+2?

and then multiple numerator and denominator by x+2

[amber wedge]

yes

[keen seal]

I tried that didnt work

wait ill write it

[amber wedge]

I can show it to you if you're not sure what to do next

[keen seal]

This is what I end up with

[keen seal]

its probably just me messing up somewhere here

[amber wedge]

It’s correct

You will need to solve the equation f(f(x))=f(x)

[keen seal]

So x/x+2 = x/3x+4?

[keen seal]

I’m not too sure what to do after this 😅

[amber wedge]

Don’t change it to ?-?=0

Just ?=?

The numerator is the same

[keen seal]

Ahh I’ll try that

[amber wedge]

The denominator should be the same

[keen seal]

Ah okay i got it

[keen seal]

Tysm!

.close

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[shrewd sapphire]

am I plussing the fractions correctly

hu

[timid sierra]

.

[fossil abyss]

Looks fine to me

You can even factorise (3k+1) from the numerator at the top

[shrewd sapphire]

I'm supposed to get k-2/10(3k+4)

[timid sierra]

sry posted to wrong channel by mistake

[shrewd sapphire]

that's not happening

[shrewd sapphire]

is ok

[shrewd sapphire]

ye

I have to dip now but I'll try again later

.close

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[brave steeple]

jst need a quick explaination, why did the limits change?

[brave steeple]

or better, HOW did they change?

the entire problem is this

plz ping

[dense idol]

substitution u=tanx, so you need to change the bounds with it

[brave steeple]

so since im using u as tanx i have to sub x with the limits?

[dense idol]

yes so you get u=tan(pi/4)=1 and u=tan(0)=0

[brave steeple]

what if i had secx

[brave steeple]

yeah i get that part, that was one of my guesses

so im guessin that works for any trig then right?

[midnight plank]

@brave steeple Has your question been resolved?

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[last slate]

Is the descriptions correct?

and is the following fulfilled?
□ All Roots (labeled on graph and values)
□ Y-intercept (labeled on graph and value)
□ Relative Maximum and Minimums (labeled on graph and values)

[midnight plank]

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[outer pasture]

help I don't know what 2 +2

[dense idol]

🗿

[slender walrus]

2+2 = <@&268886789983436800>

[velvet oar]

man I guess I've been getting that wrong my whole life

[pearl idol]

ramonov I'm not 4

are you calling me too young to be here?!

[pearl idol]

please don't troll

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[queen thistle]

((𝑛/𝑝𝑟𝑖𝑜𝑟𝑎𝑝𝑝𝑟𝑜𝑥)+𝑝𝑟𝑖𝑜𝑟𝑎𝑝𝑝𝑟𝑜𝑥)/
2

[queen thistle]

So basically

I understand the idea of this equation

N is a number (16 for example)

and priorapprox is the guess (lets say 3)

that would help me evaluate the square root of a number

so if it were 4

16/4 + 4 is 8, divided by 2 is 4

But if the wrong value is entered

say 3

it gives me 4.16666667

I'm unsure how to interpret that

[blissful pier]

so you're saying to approximate some number $\sqrt{n_0}$ you can use $t_{n}=\frac{\frac{n_0}{t_{n-1}}-t_{n-1}}{2}$?

[grand pond]

FancyBredFries

[queen thistle]

No

[blissful pier]

then what are you trying to say?

[queen thistle]

This probably isnt the real math equation

because its for a different class than math

I think priorapprox is probably a made up variable

I couldnt find anything about this online

but essentially

if you had the actual square root

of any number

and put it in

it would just stick itself out

but I'm just confused on how I can interpret wrong inputs

like 16 where my guess (or priorapprox) is 3

[blissful pier]

that's literally this ^

[queen thistle]

Okay good

[blissful pier]

let's say $\sqrt{2}\approx1.4$ then $t_{1}=\frac{\frac{2}{1.4}-1.4}{2}$

[grand pond]

FancyBredFries

[blissful pier]

at least according to the formula

,calc (2/1.4+1.4)/2

[queen thistle]

I saw the n-1's and thought it was som else

[velvet oar]

+, not -

[blissful pier]

[velvet oar]

according to OP's equation

[blissful pier]

oh! thank you, knew something was up when I got 0.014

[grand pond]

Result:

1.4142857142857

[blissful pier]

therreee we go

wow that's fast

[queen thistle]

So it didnt shoot out the same number

so we know that's not the correct sqrt of 2

[blissful pier]

but it shot out a number closer to actual sqrt(2)

,calc sqrt(2)

[grand pond]

Result:

1.4142135623731

[queen thistle]

Ohhhhh

[blissful pier]

look how close that is :) 4 digits at 1st approximation

,calc (2/1.4142857142857+1.4142857142857)/2

[grand pond]

Result:

1.4142135642136

[queen thistle]

I see

[blissful pier]

then we get 7 at the 2nd approximation, that should be enough digits to be satisfactory, but you can always compute more :)

I believe this formula comes from the Taylor approximation of sqrt(x)

but don't count me on that, there are lots of recusrive formulae approximating sqrt(x)

[queen thistle]

Okay yeah I get it now that's cool

[blissful pier]

[queen thistle]

Thank you very much

Have a good one

.close

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[blissful pier]

[midnight plank]

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[last slate]

arctg(1/3)+arctg(1/5)+arctg(1/6)+arctg(1/n) = pi/4

How do i solve it using complex numbers?

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

Can someone assist me with this problem?